# Inessential discriminant divisor

I have a question, reading a proof at some point there is written that the only inessential discriminant divisor of a cubic number field can only be $$1$$ and $$2$$, but i don't get why. If i'm not wrong the definition of inessential discriminant divisor is the follow: Given an algebraic number field $$K$$ of discriminant $$\delta_K$$, denoting with $$\mathcal{O}_K$$ his ring of integers, then one has for any $$\alpha \in \mathcal{O}_K$$ that $$\operatorname{disc}(\alpha)= [\mathcal{O}_K : \mathbb{Z}[\alpha] ]^2 \delta_K$$, then it is natural to ask for which prime $$p$$ there exsits an element $$\alpha$$ such that $$p \nmid [\mathcal{O}_K : \mathbb{Z}[\alpha] ]$$ in order to apply Kummer-Dedekind Theorem for find prime factorization of $$p \mathcal{O}_K$$. So we say for a prime $$p$$ that is an inessential discriminant divisor if for any choice of $$\alpha \in \mathcal{O}_K$$ (but maybe for any $$\alpha \in K$$, not sure) we have always that $$p \mid [\mathcal{O}_K : \mathbb{Z}[\alpha] ]$$. Then a prime is not an inessential discriminant divisor if we can always find an element $$\alpha$$ such that $$p \not\mid [ \mathcal{O}_K : \mathbb{Z}[\alpha]]$$.

Now in the case that the field is monogenic then it is trivially true that for any prime we have that we can find always $$\alpha \in \mathcal{O}_K$$ such that $$\mathcal{O}_K = \mathbb{Z}[\alpha]$$, hence $$p \not\mid 1$$. We may suppose then that the field is non monogenic, Dedekind gave an example for which a cubic number field is not monogenic and for which $$2$$ is a inessential discriminant divisor. But for $$3$$ ? How to prove that $$3$$ cannot be a inessential discriminant divisor?

I proved that if $$\alpha \in \mathcal{O}_K$$ is an algebraic integer of degree $$3$$, and let $$P(x)$$ be its minimal polynomial, if we suppose that $$P$$ is Eisenstein with respect to the prime $$p$$ then one has $$p \not\mid \left| \mathcal{O}_K/\mathbb{Z}[\alpha] \right|$$, maybe can help.

That $$p \nmid [\mathcal O_K:\mathbf Z[\alpha]]$$ when $$\alpha$$ has a minimal polynomial in $$\mathbf Z[x]$$ that is Eisenstein at $$p$$ doesn't help much, because such $$p$$ are a restricted type of ramified prime: $$(p) = \mathfrak p^n$$ where $$n = [K:\mathbf Q]$$ (a totally ramified prime in $$K$$). This does not address the more typical type of ramified prime in a cubic field where $$(p) = \mathfrak p\mathfrak q^2$$.
For a number field $$K$$, let $$i(K)$$ be the gcd of all the indices $$[\mathcal O_K:\mathbf Z[\alpha]]$$ for $$\alpha \in \mathcal O_K$$ such that $$K = \mathbf Q(\alpha)$$. It turns out that the prime factors of $$i(K)$$ are all small: they must be less than $$[K:\mathbf Q]$$: see Theorem 5.4 here.